Here you get the CBSE Class 10 Mathematics chapter 3, Pair of Linear Equations in Two Variables: NCERT Exemplar Problems and Solutions (Part-I). This part of the chapter includes solutions for Exercise 3.1 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Pair of Linear Equations in Two Variables. This exercise comprises of only the Multiple Choice Questions (MCQs) framed from various important topics in the chapter. Each question is provided with a detailed solution.

NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Mathematics Board Examination 2017-2018 as well as other competitive exams.

**Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Pair of Linear Equations in Two Variables:**

**Exercise 3.1**

**Multiple Choice Questions (MCQs)**

**Quesntion1. **Graphically, the pair of equations

6*x **-** *3*y* + 10 = 0

2*x **-** y + *9 = 0

represents two lines which are

(a) intersecting at exactly one point

(b) intersecting exactly two points

(c) coincident

(d) parallel

**Solution: **(d)

**Explanation:**

On comparing with *a*_{1}*x *+ *b*_{1}*y *+ *c*_{1} = 0 and *a*_{2}*x *+ *b*_{2}*y *+ *c*_{2} = 0 we have *a*_{1}* *= 6, *b*_{1}* *= ‒3, *c*_{1} = 10 and *a*_{2 }= 2, *b*_{2}*y *= ‒1, *c*_{2} = 9.

So, the given system of equations have no solution, this will be possible if two lines never intersect each other.

In other words both the lines are parallel to each other.

Rough figure is shown below for reference,

**Quesntion2.**The pair of equations *x *+ 2*y* + 5 = 0 and - 3*x**-* 6*y* + 1 = 0 has

(a) a unique solution

(b) exactly two solutions

(c) infinitely many solutions

(d) no solution

**Solution: **(d)

Given, equations are *x* + 2*y* + 5 = 0 and - 3*x **-** *6*y + *1 = 0

On comparing with *a*_{1}*x *+ *b*_{1}*y *+ *c*_{1} = 0 and *a*_{2}*x *+ *b*_{2}*y *+ *c*_{2} = 0, we have

Therefore, the pair of equations has no solution.

**Quesntion3. **If a pair of linear equations is consistent, then the lines will be

(a) parallel

(b) always coincident

(c) intersecting or coincident

(d) always intersecting

**Solution: **(c)* *

If two lines are parallel then, they have no solution pair of linear equations is inconsistent;

If two lines are coincident then, they have infinite solution and pair of linear equations is consistent;

If two lines are intersecting then, they have unique solution and pair of linear equations is consistent.

**Quesntion4.**The pair of equations *y* = 0 and *y* = ‒7 ha**s**

(a) one solution

(b) two solutions

(c) infinitely many solutions

(d) no solution

**Solution: **(d)* *

If we draw a rough graph of both the lines (as shown below) we will find that both lines are parallel. So, they have no solution.

**Quesntion5. **The pair of equations *x *= *a *and *y = b *graphically represents lines which are

(a) parallel

(b) intersecting at (*b*, *a*)

(c) coincident

(d) intersecting at (*a*, *b*)

**Solution: **(d)

*x* = *a* is the equation of a straight line which is parallel to the *y*–axis lying at a distance ‘a’ from it. Also, *y *= *b* is the equation of a straight line parallel to the *x*–axis lying at a distance ‘b’ from it. Thus the two lines come to intersect at a point (*a*, *b*).

**Quesntion6.** For what value of *k*, do the equations 3*x* -*y* + 8 = 0 and 6*x* -*k y* = -16 represent coincident lines?

* *

**Quesntion7. **If the lines given b*y *3*x+ *2*ky *= 2 and 2*x* + 5*y* = 1 are parallel, then the value of *k *is

**Quesntion8. **The value of *c *for which the pair of equations *cx **-** y *= 2 and 6*x **-** *2*y = *3 will have infinitely many solutions is

(a) 3

(b) -3

(c) -12

(d) no value

**Solution: **(d)* *

Condition for two lines *a*_{1}*x *+ *b*_{1}*y *+ *c*_{1} = 0 and *a*_{2}*x *+ *b*_{2}*y *+ *c*_{2} = 0 to have infinitely many solutions is:

As, *c* has different values, therefore, there will be no value of *c* for which the pair of equations will have infinitely many solutions.

**Quesntion9. **One equation of a pair of dependent linear equations is - 5*x *+ 7*y *- 2 = 0. The second equation can be

(a) 10*x + *14*y + *4 = 0

(b) 10*x* -14*y* + 4 = 0

(c) 10*x* + 14*y* + 4=0

(d) 10*x *-14*y* + 4=0

**Solution: **(d)* *

Condition for two linear equations *a*_{1}*x *+ *b*_{1}*y *+ *c*_{1} = 0 and *a*_{2}*x *+ *b*_{2}*y *+ *c*_{2} = 0 to be dependent is:

** **

**Quesntion10. **A pair of linear equations which has a unique solution** ***x* = 2 and *y* = - 3 is

(a)* x + y =* 1 and 2*x* + 3*y* = - 5

(b) 2*x *+ *y* = - 11 and 4*x* + 10*y* = - 22

(c) 2*x **-** y = *1 and 3*x *- 2*y* = 0

(d) *x **-** *4*y* - 14 = 0 and 5*x **-** y **-** *13 = 0

**Solution: **(b)* *

If *x* = 2, *y* = ‒3 is a unique solution of a pair of equations, then these values will satisfy both pair of equations.

So, we will check by putting *x* = 2, *y* = ‒3 in all options and check which one of the four options is correct.

In option (b),

Putting *x* = 2, *y* = ‒3 in LHS of equation, 2*x *+ *y* = - 11

⇒ 2(2) + 5(-3) = 4 - 15 = 11 = RHS

Putting *x* = 2, *y* = ‒3 in LHS of equation 4*x*+ 10*y* = ‒22

⇒4 (2) + 10(-3) = 8 - 30 = -22 = RHS

**Quesntion11. **If *x = a *and *y = b *is the solution of the equations *x **-** y = *2* *and *x *+ *y* = 4, then the values of *a *and *b *are, respectively

(a) 3 and 5

(b) 5 and 3

(c) 3 and 1

(d) 1 and -3

**Solution: **(c)

As, *x* = *a* and *y* = *b *being the solution of the equations *x **-** y* = 2 and *x* + *y* = 4, must satisfy both the equations.

Thus, on putting *x* = *a* and *y* = *b* in both the equations, we get:

* a **- b* = 2 …(i)

And *a + b *= 4 ...(ii)

On solving equations (i) and (ii), we get 2*a* = 6 or *a* = 3.

⟹ *a *= 3 and *b* = 1.

**Quesntion12. **Aruna has only Rs. 1 and Rs. 2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is Rs. 75, then the numbers of Rs. 1 and Rs.2 coins are, respectively

(a) 35 and 15

(b) 35 and 20

(c) 15 and 35

(d) 25 and 25

**Solution: **(d)

Let total number of Rs. 1 coins = *x*** **

And total number of Rs. 2 coins = *y*

Now, according to the conditions given in question *x* + *y* = 50 …(i)

Also, *x *× 1+ *y×*2=75** **

Þ *x *+ 2*y* = 75 ...(ii)

Solving (i) and (ii)

(*x* + 2*y*) - (*x* + *y*) = 75 - 50

Þ *y* = 25

As, *x* + *y* =50** **

Therefore, for *y* = 25, we get *x* = 25.

**Quesntion13. **The father's age is six times his son's age. Four years hence, the age of the father will be four times his son's age. The present ages (in year) of the son and the father are, respectively

(a) 4 and 24

(b) 5 and 30

(c) 6 and 36

(d) 3 and 24

**Solution: **(c)

Let present age of father = *x*

And present age of son = *y *

It is given in question that, *x =* 6*y *…(i)

According to the question,

(*x *+* *4) = 4 (*y* + 4)

⇒ *x *- 4*y* = 12 ...(ii)

Using equations (i) and (ii), we get:

6*y *- 4*y *= 12** **

Þ 2*y* = 12** **

Þ *y *= 6

If *y = *6, then *x *= 36

Therefore, present age of father is 36 year and present age of son is 6 year.** **

**CBSE Class 10 Mathematics Syllabus 2017-2018**

**CBSE Class 10 NCERT Textbooks & NCERT Solutions**

**NCERT Solutions for CBSE Class 10 Maths**

**NCERT Exemplar Problems and Solutions Class 10 Science: All Chapters**

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